Mathematics - Functions Question with Solution | TestHub

(iii) Given that $y(x)=\sqrt{e^{{\cos }^{-1}\left({\log }_4{x}^2\right)}}$.

The function is defined if (i) $x^2>0$ which is true for all x (ii) $-1\le {\log }_4{x}^2\le 1$.

Now, (ii) $\Rightarrow 4^{-1}\le x^2\le 4\Rightarrow \frac{1}{4}\le x^2\le 4\Rightarrow x\in \left[\frac{1}{2},2\right]$ or $x\in \left[-2,-\frac{1}{2}\right]$.

Hence the domain of the function is $\left[-2,-\frac{1}{2}\right]\cup \left[\frac{1}{2},2\right]$.

To find out the range, let $y_1={\log }_4{x}^2$ so that $y=\sqrt{e^{{\cos }^{-1}\left(y_1\right)}}$.

Again, let $y_2={\cos }^{-1}\left(y_1\right)$. $\therefore y=\sqrt{e^{y_2}}$ where $y_2={\cos }^{-1}\left(y_1\right)$ and $y_1={\log }_4{x}^2$.

Now, for $x=\frac{1}{2}($ or $-\frac{1}{2})y_1={\log }_4\left(\frac{1}{4}\right)={\log }_4\left(4^{-1}\right)=-1$ And for $\mathrm{x}=2$ (or -2 ), $y_1={\log }_4(4)=1$. Hence $y_1$ lies between -1 and 1 i.e. $-1\le y_1\le 1\Rightarrow {\cos }^{-1}(-1)\ge {\cos }^{-1}\left(y_1\right)\ge {\cos }^{-1}(1)$ $\Rightarrow \pi \ge y_2\ge 0\Rightarrow 0\le y_2<\pi$.

Again $0\le y_2\le \pi \Rightarrow e^{y_2}\le e^{\pi }\Rightarrow 1\le e^{y_2}\le e^{\pi }\Rightarrow 1\le \sqrt{e^{y_2}}\le \sqrt{e^{\pi }}\Rightarrow 1\le y\le \sqrt{e^{\pi }}$.

Hence the range of the function is $\left[1,\sqrt{e^{\pi }}\right]$.